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If differentials are just an increment ratio of x and y along a tangent with slope f'(x), why are differentials thought of as being "incredibly" (or if you want to get murdered) "infinitesimally" small? Does it not remain true that on a tangent with slope f'(x), that dy/dx = Δy/Δx is the same ratio for all increments of x?

It was my understanding that the only reason we are able to say that dy/dx = Δy/Δx = f'(x) is because if y is a linear function of x, then the limit as Δx approaches 0 for the increment ratio Δy/Δx is the same as the ratio Δy/Δx when the limit is not computed. Or

dy/dx = limit Δx→0 Δy/Δx = Δy/Δx = f'(x), or simply dy/dx = Δy/Δx = f'(x) (when y is a linear function of x).

So why is there the unspoken convention that dy and dx are very small increments?

Ethan Penrod
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2 Answers2

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If $y=f(x)$ and $y+\delta y=f(x+\delta x)$, the derivative is
$$f'(x)=\frac{dy}{dx}=\lim_{\delta x\to 0} \frac{\delta y}{\delta x}\text{ .} $$ The expression $\frac{dy}{dx}$ is not actually a fraction, although it looks like one.

If you think of $\frac{dy}{dx}$ as a fraction, then $dy$ and $dx$ would be the limiting, very small values of $\delta y$ and $\delta x$. However the limiting values of both $\delta y$ and $\delta x$ are zero, so the fraction would be $\frac{0}{0}$, which is undefined.

I have seen $dx$ and $dy$ used by Physics lecturers. It needs to be understood that $dx$ and $dy$ are small enough that their ratio is close enough to $\frac{dy}{dx}$ for the purposes of the lecture. A mathematically correct derivation would not have been within the scope of the Physics lecture.

Peter
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  • "The expression dy/dx is not actually a fraction, although it looks like one" that's originally how the Leibniz notation for derivatives was introduced in my text book. It was made clear that dy/dx was to be thought of as a symbol, rather than a quotient of two values (dy and dx). That interpretation is in stark contrast to the chapter on differentials where the author explicitly states "The new feature of dy/dx = f '(x) in our present discussion is that the Leibniz symbol on the left not only looks like a fraction, but is a fraction: dy = differential of y / dx = differential of x – Ethan Penrod Jun 27 '20 at 20:56
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The two places where elementary calculus uses a notation involving differentials are the derivative $dy/dx$ and the integral $\int_a^b f(x)\,dx$. Both of these are defined by a limiting process, i.e., they are approximated by quantities involving small increments of $y$ and $x$. In the case of the derivative, the approximants are difference quotients $\Delta y/\Delta x$, and in the case of the integral, the approximants are Riemann (or Darboux) sums. In both cases, the approximation gets better as the increments get smaller. So it is natural to think of the differentials as representing increments in the context where they are very small --- and the smaller the better.

Concerning the first paragraph of your question: It is certainly true that $f'(x)=dy/dx$ is the slope of the whole tangent line to the graph of $f$ at the point $(x,f(x))$. You can take any (non-zero) increment $\Delta x$ and the corresponding increment $\Delta y$ along the tangent line and their ratio will be $dy/dx$. But the intention behind the derivative is to take $\Delta y$ along the graph of $f$; after all $y$ is supposed to represent $f(x)$ and so $\Delta y$ should represent the change in $f(x)$ caused by the increment $\Delta x$ of $x$. And this intended slope $\Delta y/\Delta x$ will not coincide with $dy/dx$ (unless $f$ is linear); it will only be approximately $dy/dx$ when $\Delta x$ is small --- the smaller the better. So again, $dx$ and $dy$ are behaving like incredibly small $\Delta x$ and $\Delta y$.

Andreas Blass
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